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---
title: Inner Product Spaces
parent: Functional Analysis Basics
nav_order: 6
has_children: true
has_toc: false
---
# {{ page.title }}
{% definition Inner Product Space %}
An *inner product* (or *scalar product*) on a real or complex vector space $X$
is a mapping
$$
\innerp{\cdot}{\cdot} : X \times X \to \KK
$$
that is
- linear in its second argument
$$
\innerp{x}{y+z} = \innerp{x}{y} + \innerp{x}{z} \qquad
\innerp{x}{\alpha y} = \alpha \innerp{x}{y}
$$
- conjugate symmetric
$$
\overline{\innerp{x}{y}} = \innerp{x}{y}
$$
- nondegenerate
An *inner product space* (or *pre-Hilbert space*) is a pair $(X,\innerp{\cdot}{\cdot})$
consisting of a real or complex vector space $X$
and an inner product $\innerp{\cdot}{\cdot}$ on $X$.
{% enddefinition %}
{% proposition Norm Induced by an Inner Product %}
If $\innerp{\cdot}{\cdot}$ is an inner product
on a real or complex vector space $X$, then
$$
\norm{x} = \sqrt{\innerp{x}{x}} \qquad \forall x \in X
$$
defines a norm on $X$.
{% endproposition %}
In this sense, every inner product space is also a normed space.
As a consequence it is also a metric space and a topological space.
The next theorem shows how the inner product can be recovered from the norm.
{% theorem * Polarization Identity %}
For all vectors $x$ and $y$ of a real inner product space
$$
4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2.
$$
For all vectors $x$ and $y$ of a complex inner product space
$$
4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2 + i \norm{x-iy}^2 - i \norm{x+iy}^2.
$$
{% endtheorem %}
Note that the complex polarization identity takes the slightly different form
$$
4 \innerp{x}{y} = \norm{x+y}^2 - \norm{x-y}^2 + i \norm{x\mathrel{\color{red}+}iy}^2 - i \norm{x\mathrel{\color{red}-}iy}^2,
$$
if we follow the convention that the inner product is conjugate linear in its second argument.
{% proof %}
In the real case, the inner product is symmetric, and we have
$$
\norm{x \pm y}^2 = \norm{x}^2 \pm 2 \innerp{x}{y} + \norm{y}^2
$$
for all vectors $x$ and $y$.
Taking the difference yields the desired result.
In the complex case, the inner product is conjugate symmetric, and we have
$$
\norm{x \pm y}^2 = \norm{x}^2 \pm 2 \Re \innerp{x}{y} + \norm{y}^2
$$
for all vectors $x$ and $y$. This implies
$$
\begin{aligned}
\norm{x + \phantom{i}y}^2 - \norm{x - \phantom{i}y}^2 &= 4 \Re \innerp{x}{y}, \\
\norm{x - iy}^2 - \norm{x + iy}^2 &= 4 \Im \innerp{x}{y}.
\end{aligned}
$$
The second equation follows from the first by
substituting $y$ with $-iy$ and
using that $\Re \innerp{x}{-iy} = \Re (-i\innerp{x}{y}) = \Im \innerp{x}{y}$.
To obtain the polarization Identity, multiply the second equation with $i$ and then add it to the first.
{% endproof %}
{% theorem * General Polarization Identity %}
Let $X$ be a complex inner product space.
Let $\zeta$ be a $n$-th root of unity with $\zeta \ne 1$ and $\zeta^2 \ne 1$.
Then
$$
\innerp{x}{y} = \frac{1}{n} \sum_{k=0}^{n-1} \zeta^k \norm{x + \zeta^k y}^2 \qquad \forall x,y \in X.
$$
{% endtheorem %}
As a special case, for $\zeta = i$ and $n=4$, we obtain
$$
\innerp{x}{y} = \frac{1}{4} \sum_{k=0}^{3} i^k \norm{x + i^k y}^2.
$$
{% proof %}
TODO
{% endproof %}
For an arbitrary normed space,
the polarization identity does not, in general,
define an inner product.
The following theorem, gives a condition for when it does.
{% theorem * Parallelogram Law %}
Let $X$ be a real or complex normed space.
A norm $\norm{\cdot}$ on $X$ is induced by
an inner product $\innerp{\cdot}{\cdot}$ on $X$,
if and only if $\norm{\cdot}$ satisfies the *parallelogram law*
$$
\norm{x+y}^2 + \norm{x-y}^2 = 2 \norm{x}^2 + 2 \norm{y}^2 \qquad \forall x,y \in X.
$$
In this case, the inner product is uniquely determined by $\norm{\cdot}$
and given by the polarization identity.
{% endtheorem %}
{% theorem * Stewart’s Theorem %}
Let $x$, $y$, $z$ be vectors of an inner product space.
If $x$, $y$ and $z$ are colinear and $y$ lies inbetween $x$ and $y$,
then we have
$$
\norm{p-x}^2 \norm{y-z} + \norm{p-z}^2 \norm{x-y} =
\big\lparen \norm{p-y}^2 + \norm{x-y} \norm{y-z} \big\rparen \norm{x-z}
$$
{% endtheorem %}
---
{% theorem * Cauchy–Schwarz Inequality %}
For all vectors $x$ and $y$ of an inner product space
(with inner product $\innerp{\cdot}{\cdot}$ and induced norm $\norm{\cdot}$)
$$
\abs{\innerp{x}{y}} \le \norm{x} \norm{y},
$$
and equality holds precisely when $x$ and $y$ are linearly dependent.
{% endtheorem %}
Expressed only in terms of the inner product, the Cauchy–Schwarz Inequality reads
$$
\abs{\innerp{x}{y}}^2 \le \innerp{x}{x} \innerp{y}{y}.
$$
{% proof %}
TODO
{% endproof %}
{% corollary Continuity of the Inner Product %}
The inner product is jointly norm continuous.
{% endcorollary %}
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